Speed, Time and Distance
- A man covers 20 kms in 2 hours.
Find the distance covered by him in 5 1 hours. 2
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Distance = 20 kms
Time = 2 hours∴ Speed = Distance Time = 20 = 10 km per hr. 2
Now, we have, Speed = 10 km per hr.Time = 11 hr. 2
∴ Distance = Speed × Time= 10 × 11 = 55 km 2 Correct Option: C
Distance = 20 kms
Time = 2 hours∴ Speed = Distance Time = 20 = 10 km per hr. 2
Now, we have, Speed = 10 km per hr.Time = 11 hr. 2
∴ Distance = Speed × Time= 10 × 11 = 55 km 2
- Express speed of 60 metres per sec. in km per hour.
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60 metres per sec.
= 60 × 18 km per hr. 5
= 216 km per hr.Correct Option: B
60 metres per sec.
= 60 × 18 km per hr. 5
= 216 km per hr.
- Express speed of 36 km per hr. in metres per second.
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36 km/hr.
= 36 × 5 m/sec. 18
= 10 m/sec.Correct Option: A
36 km/hr.
= 36 × 5 m/sec. 18
= 10 m/sec.
- A man walking at 3 km/hour crosses a square field diagonally in 2 minutes. The area of the field (in square metre) is
-
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Speed of person = 3 kmph
= 3000 m./min. 60
= 50 m./min.
∴ Length of the diagonal of square field
= 50 × 2 = 100 metre∴ Required time = 1 × (100)2 2
= 5000 sq. metreCorrect Option: B
Speed of person = 3 kmph
= 3000 m./min. 60
= 50 m./min.
∴ Length of the diagonal of square field
= 50 × 2 = 100 metre∴ Required time = 1 × (100)2 2
= 5000 sq. metre
- A and B are 15 kms apart and when travelling towards each other meet after half an hour whereas they meet two and a half hours later if they travel in the same direction. The faster of the two travels at the speed of
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A’s speed = x kmph.
B’s speed = y kmph.
When A and B move in opposite
directions they meet at C and
when they move in the same direction, they meet at D.
Case I,
AC + CB = ABx + y = 15 2 2
⇒ x + y = 30 ..... (i)
Case II,
AD – BD = AB⇒ x × 5 − y × 5 = 15 2 2 5 (x – y) = 15 2 ⇒ x – y = 15 × 2 = 6 .....(ii) 5
∴ On adding equations (i) and (ii),
x + y + x – y = 30 + 6
⇒ 2x = 36⇒ x = 36 = 18 kmph. 2 Correct Option: B
A’s speed = x kmph.
B’s speed = y kmph.
When A and B move in opposite
directions they meet at C and
when they move in the same direction, they meet at D.
Case I,
AC + CB = ABx + y = 15 2 2
⇒ x + y = 30 ..... (i)
Case II,
AD – BD = AB⇒ x × 5 − y × 5 = 15 2 2 5 (x – y) = 15 2 ⇒ x – y = 15 × 2 = 6 .....(ii) 5
∴ On adding equations (i) and (ii),
x + y + x – y = 30 + 6
⇒ 2x = 36⇒ x = 36 = 18 kmph. 2